Solving for the Inverse of Matrix A

[t_n_p]

Homework Statement

Give the following matrix A = [2 1; 3 2] show that (A^2)-4A+I=0 where I is the 2x2 identity matrix. Hence use your result to deduce the inverse of A.

The attempt at a solution

I can do the show part no problem, but I'm stuck on how to find inverse of A. I multiply both sides of the equation by inverse A and get..

(A^-1)(A^2) - 4A(A^-1) + I(A^-1) = 0(A^-1)

Can you apply index laws on matrices? Any suggestions on what I should do next? Thanks!

 

[Dick]

The equation you just wrote simplifies to A-4I+A^(-1)=0. Do you see how? Can you solve that for A^(-1)? You were practically done!

 

[t_n_p]

yeah i see now, was just unsure whether I could apply index law for the first set of terms. just to refresh my memory!

So any matrix multiplied by its inverse gives the identity matrix of the same dimensions and

any matrix multiplied by the identity matrix gives the original matrix

 

[Dick]

What index law? You are just using A*A^(-1)=I. You mean A^n*A^m=A^(n+m)? Sure you can.

 

[t_n_p]

Yeah I just wanted to know if you could use A^n*A^m=A^(n+m).

Now everything is ok!

 

[HallsofIvy]

It's not really a matter of exponents (not "indices"), just the distributive law. You already know that (A^2)-4A+I=0 so obviously 4A- A^2= A(4I-A^2)= I. Now use the definition of "multiplicative inverse".

 

[ZioX]

Note that A does not have an inverse iff the characteristic polynomial has zero as a root iff the constant term of the characteristic polynomial is zero.

 

[HallsofIvy]

Yes, but the fact that (A^2)-4A+I=0 shows that neither of those is true!